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p_n(x) = 1/2 ((x+\sqrt{x^2-1} )^n+ (x-\sqrt{x^2-1} )^n

Source: 1996 Swedish Mathematical Competition p3

April 2, 2021
algebraradicalSequence

Problem Statement

For every positive integer nn, we define the function pnp_n for x≥1x\ge 1 by pn(x)=12((x+x2−1)n+(x−x2−1)n).p_n(x) = \frac12 \left(\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n\right). Prove that pn(x)≥1p_n(x) \ge 1 and that pmn(x)=pm(pn(x))p_{mn}(x) = p_m(p_n(x)).
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